Exam2_2 9 - db dt = 2 h ‡ 5-2 A h · cms/min ....

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Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 9 When x = 100, therefore, marginal cost = C 0 (100) = $20 per pair . keywords: cost, marginal cost, word problem, derivative 018 (part 1 of 1) 10 points The height of a triangle is increasing at a rate of 2 cm/min while its area is increasing at a rate of 5 sq. cms/min. At what speed is the base of the triangle changing when the height of the triangle is 5 cms and its area is 25 sq. cms? 1. speed = 9 5 cms/min 2. speed = 7 5 cms/min 3. speed = 11 5 cms/min 4. speed = 8 5 cms/min 5. speed = 2 cms/min correct Explanation: Let b be the length of the base and h the height of the triangle. Then the triangle has area = A = 1 2 b h . Thus by the Product Rule, dA dt = 1 2 b dh dt + h db dt · , and so db dt = 1 h 2 dA dt - b dh dt · = 2 h dA dt - A h dh dt · , since b = 2 A/h . Thus, when dh dt = 2 , and dA dt = 5 , we see that
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Unformatted text preview: db dt = 2 h ‡ 5-2 A h · cms/min . Consequently, at the moment when h = 5 and A = 25 , the base length is changing at a speed = 2 cms/min . keywords: speed, related rates 019 (part 1 of 1) 10 points A 15 foot ladder is leaning against a wall. If the foot of the ladder is sliding away from the wall at a rate of 6 ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 12 feet away from the base of the wall? 1. speed = 23 3 ft/sec 2. speed = 22 3 ft/sec 3. speed = 8 ft/sec correct 4. speed = 7 ft/sec 5. speed = 20 3 ft/sec Explanation: Let y be the height of the ladder when the foot of the ladder is x feet from the base of the wall as shown in Fgure...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas.

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